Hy,
我有以下代码:
public AppThread(SocketChannel socketChannel){
this.socketChannel=socketChannel;
}
public void run(){
try{
ByteBuffer bb = ByteBuffer.allocate(11);
socketChannel.read(bb);
//byte[] b = new byte[bb.capacity()];
// bb.get(b, 0, 11);
System.out.println(bb.toString());
byte[] a = new byte[11];
CharBuffer cb = bb.asCharBuffer();
System.out.println(cb);
bb.get(a);
App app=new App();
// String an = new String(b);
//String zodie = Zodie.getZodie(an);
//b = new byte[zodie.length()];
//b = zodie.getBytes();
bb.clear();
//bb.put(b);
socketChannel.write(bb);
socketChannel.close();
}
catch(IOException e){
System.err.println("Server comunication error : "+e.getMessage());
}
}
}
和接收字符串并返回字符串的zodie静态方法。
如何将写入SocketChannel的字符串作为参数传递给zodiac静态方法。
我提到在客户端我发送一个字节数组,我已经检查过了,没关系。
客户方:
byte[] a = an.getBytes();
System.out.println(new String(a));
ByteBuffer bb=ByteBuffer.allocate(11);
// Varianta 1
bb.put(a);
// Varianta 2
// LongBuffer lb=bb.asLongBuffer();
// lb.put(0,m).put(1,n);
try{
sc.write(bb);
bb.clear();
sc.read(bb);
// Varianta 1
//a = new byte[bb.remaining()];
zodie=bb.toString();
// Varianta 2
// r=lb.get(0);
System.out.println("Zodia : "+ zodie);
sc.close();
此致
我收到服务器错误:
Server ready...
java.nio.HeapByteBuffer[pos=1 lim=11 cap=11]
Exception in thread "pool-1-thread-1" java.nio.BufferUnderflowException
at java.nio.HeapByteBuffer.get(Unknown Source)
at java.nio.ByteBuffer.get(Unknown Source)
at server.AppThread.run(AppThread.java:27)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)
答案 0 :(得分:0)
问题出在客户端。我使用了一个charbuffer作为一个包装器
Scanner scanner=new Scanner(System.in);
String m;
System.out.println("Enter the date (yyyy-mm-dd) :");
m=scanner.next()+" ";
CharBuffer c = CharBuffer.wrap(m);
System.out.println("Sending date ...: " + c);
ByteBuffer b = Charset.forName("ISO-8859-1").encode(c);
b.compact();
System.out.println("Bytebuffer has the capacity of "+
b.capacity() + "pointer position on bytebuffer is on: "+ b.position() + " and the limit is:" + b.limit());
b.flip();
在服务器端:
CharBuffer c;
ByteBuffer bb = ByteBuffer.allocate(11);
System.out.println("Server allocated a number of 11 octets to ByteBuffer");
socketChannel.read(bb);
bb.flip();//sets the Position to 0 and limit to the number of bytes to be read.
CharBuffer c = Charset.forName("ISO-8859-1").decode(bb);
System.out.println("Got " + c);
byte[] byteArray = new byte[11];
bb.get(byteArray);
System.out.println("Server got from client the string: " +new String(byteArray));
String an = new String(byteArray);
bb.clear();
答案 1 :(得分:-1)
你的问题是在行bb.get(a);
您指定11个字符数组作为参数,因此方法get()要求bb包含11个字符,但它只包含1个。
原因是当你向套接字写入11个字符时,它们不能保证一次到达。他们可以分成小组,如1和10或5和6等。
您需要查看并读取套接字并检查read()调用的结果。一些事情:
for (;;) {
int numberOfBytesRead =socket.read(buf);
if (numberOfBytesRead < 0) { break; }
}